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  <title>利用递归求1~n的阶乘</title>
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<body>
  <script>
    // 利用递归函数求1~n的阶乘 1 * 2 * 3 * 4 * ..n
    function fn(n) {
      if (n === 1) {
        return 1;
      }
      return n * fn(n - 1);
    }
    console.log(fn(3));
    console.log(fn(4));
    // 详细思路 假如用户输入的是3
    //return  3 * fn(2)
    //return  3 * (2 * fn(1))
    //return  3 * (2 * 1)
    //return  3 * (2)
    //return  6
  </script>
</body>
</html>